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Sergeeva-Olga [200]
3 years ago
11

From a sample with nequals8​, the mean number of televisions per household is 4 with a standard deviation of 1 television. Using

​ Chebychev's Theorem, determine at least how many of the households have between 2 and 6 televisions.
Mathematics
1 answer:
OlgaM077 [116]3 years ago
3 0

Answer:

75% of the households have between 2 and 6 televisions

Step-by-step explanation:

From the question, we can deduce the following;

sample size n= 8

sample mean μ = 4

standard deviation σ = 1

Using Chebychev’s theorem;

P(2 ≤ X ≤ 6) = P(2-4 ≤ (X - μ) ≤ 6-4)

= P(-2 ≤ (X-μ) ≤ 2) = P(|X-μ| ≤ kσ) ≥ (1 - 1/k^2) ≥ (1- 1/2^) = 1- 0.25 = 0.75 ( same as 75%)

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A researcher was interested in comparing the resting heart rate of people who exercise regularly and people who do not exercise
alekssr [168]

Answer:

t=\frac{(73.5-69.3)-0}{\sqrt{\frac{10.2^2}{16}+\frac{8.7^2}{12}}}}=1.173  

Using the critical value method we need to find a critical value in the t distribution who accumulates 0.025 of the area on the right and we got:

t_{cric}= 2.06

So then since our calculated value is lower then the critical value we fail to reject the null hypothesis and we can't conclude that the true mean resting pulse rate of people who do not exercise regularly is larger than the mean resting pulse rate of people who exercise regularly.

Step-by-step explanation:

Information given

\bar X_{1}=73.5 represent the mean for sample of people who do not exercise

\bar X_{2}=69.3 represent the mean for sample of people who do exercise

s_{1}=10.2 represent the sample standard deviation for 1  

s_{2}=8.7 represent the sample standard deviation for 2  

n_{1}=16 sample size for the group 1

n_{2}=12 sample size for the group 2

\alpha=0.01 Significance level

t would represent the statistic

System of hypothesis

We want to verify if the mean resting pulse rate of people who do not exercise regularly is larger than the mean resting pulse rate of people who exercise regularly, the system of hypothesis would be:  

Null hypothesis:\mu_{1}-\mu_{2} \leq 0  

Alternative hypothesis:\mu_{1} - \mu_{2}> 0  

The statistic for this case would be given by:

t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}} (1)  

The degrees of freedom are given by:

df=n_1 +n_2 -2=16+12-2=26  

Replacing the info given we got:

t=\frac{(73.5-69.3)-0}{\sqrt{\frac{10.2^2}{16}+\frac{8.7^2}{12}}}}=1.173  

Using the critical value method we need to find a critical value in the t distribution who accumulates 0.025 of the area on the right and we got:

t_{cric}= 2.06

So then since our calculated value is lower then the critical value we fail to reject the null hypothesis and we can't conclude that the true mean resting pulse rate of people who do not exercise regularly is larger than the mean resting pulse rate of people who exercise regularly.

6 0
3 years ago
An example of financial responsibility is
Maurinko [17]

Answer:

keeping financial records organized

4 0
3 years ago
Find the area of the trapezoid with a base of 19 inches, height 12.6 inches and base 29.2 inches
aksik [14]
Minor base: b=19 inches
Height: h=12.6 inches
Major base: B=29.2 inches

Area of the trapezoid: A
A=(b+B)h/2
Replacing the values:
A=(19 inches + 29.2 inches) (12.6 inches) / 2
A=(48.2 inches) (12.6 inches) / 2
A= (607.32 inches^2 ) /2
A= 303.66 inches^2

Answer: The area of the trapezoid is 303.66 square inches
3 0
3 years ago
Read 2 more answers
In a class of 20 students, 9 are female and 15 have an A in the class. There are 2 students who are male and do not have an A in
Y_Kistochka [10]

Answer:

3/5

Step-by-step explanation:

9/15=3/5

since 15 had an a you divide the girls over the people who had an A

3 0
3 years ago
C.<br> In (MN) = In M + In N, for M&gt; 0 and N&gt; 0<br> True or false
melisa1 [442]
That statement is true
8 0
3 years ago
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